Let $(x_1,y_1)(x_2,y_2)... (x_n,y_n)$ be independent pairs such that $y_i=\theta x_i+\epsilon_i$ where $x_i$ and $\epsilon_i$ are iid Normal (0,1) for $i=1,2..n$

It was previously computed that

$f(x,y)=\frac{1}{2\pi}e^{\frac{-1}{2}x^2-\frac12(y-\theta x)^2 }$

And that the MLE for $\theta$ is $\theta^*=\frac{\sum_{i=1}^ny_i}{\sum_{i=1}^nx_i}$ which is unbiased for $\theta$

I need to know if this estimator achieves the CRLB.

Here is what I have so far:

For any estimator W(M) of $\theta$ using sample $M_i's$ of$M$ which are iid $i=1..n$, the CRLB is given by

$\frac{[\frac{d}{d\theta}E(W(M))]^2}{nE_\theta[(\frac{d}{d\theta}lnf(M|\theta)]^2}$

So the numerator is 1 since $E(W(M))=\theta$.

For the denominator, I worked it out as $n(\theta^2(2-\theta^2))$.

Is this correct? It seems iffy coz it can have negative values.

Another problem is in computing the variance of the estimator.

$Var(\theta^*)=Var(\frac{\sum_{i=1}^ny_i}{\sum_{i=1}^nx_i}$). How do I go about this?

Let $(x_1,y_1)(x_2,y_2)... (x_n,y_n)$ be independent pairs such that $y_i=\theta x_i+\epsilon_i$ where $x_i$ and $\epsilon_i$ are iid Normal (0,1) for $i=1,2..n$

It was previously computed that

$f(x,y)=\frac{1}{2\pi}e^{\frac{-1}{2}x^2-\frac12(y-\theta x)^2 }$

And that the MLE for $\theta$ is $\theta^*=\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}$ which is unbiased for $\theta$

I need to know if this estimator achieves the CRLB.

Here is what I have so far:

For any estimator W(M) of $\theta$ using sample $M_i's$ of$M$ which are iid $i=1..n$, the CRLB is given by

$\frac{[\frac{d}{d\theta}E(W(M))]^2}{nE_\theta[(\frac{d}{d\theta}lnf(M|\theta)]^2}$

So the numerator is 1 since $E(W(M))=\theta$.

For the denominator, I worked it out as $n(\theta^2(2-\theta^2))$.

Is this correct? It seems iffy coz it can have negative values.

Another problem is in computing the variance of the estimator.

$Var(\theta^*)=Var(\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}$). How do I go about this?